To solve for \( \int_2^5 f(x) \, dx \) given that \( \int_0^5 f(x) \, dx = 3 \) and \( \int_0^2 f(x) \, dx = -2 \), follow these steps: 1. **Understand the relationship between the integrals**: The integral from 0 to 5 can be split into two parts: \[ \int_0^5 f(x) \, dx = \int_0^2 f(x) \, dx + \int_2^5 f(x) \, dx \] 2. **Substitute the given values**: \[ 3 = \int_0^2 f(x) \, dx + \int_2^5 f(x) \, dx \] Given that \( \int_0^2 f(x) \, dx = -2 \), substitute this value into the equation: \[ 3 = -2 + \int_2^5 f(x) \, dx \] 3. **Solve for \( \int_2^5 f(x) \, dx \)**: \[ 3 = -2 + \int_2^5 f(x) \, dx \] \[ \int_2^5 f(x) \, dx = 3 + 2 \] \[ \int_2^5 f(x) \, dx = 5 \] So, the value of \( \int_2^5 f(x) \, dx \) is \( 5 \). The correct answer is: - \( 5 \)

To find \(\int_{-1}^{1} f(x) \, dx\), we can use the given integrals and the properties of definite integrals. We know: \[ \int_{-4}^{5} f(x) \, dx = -8.0 \] \[ \int_{-4}^{-1} f(x) \, dx = -2.4 \] \[ \int_{1}^{5} f(x) \, dx = -6.1 \] We can break down the integral \(\int_{-4}^{5} f(x) \, dx\) into parts: \[ \int_{-4}^{5} f(x) \, dx = \int_{-4}^{-1} f(x) \, dx + \int_{-1}^{1} f(x) \, dx + \int_{1}^{5} f(x) \, dx \] Substitute the known values: \[ -8.0 = -2.4 + \int_{-1}^{1} f(x) \, dx + (-6.1) \] Combine the constants: \[ -8.0 = -2.4 - 6.1 + \int_{-1}^{1} f(x) \, dx \] \[ -8.0 = -8.5 + \int_{-1}^{1} f(x) \, dx \] Solve for \(\int_{-1}^{1} f(x) \, dx\): \[ \int_{-1}^{1} f(x) \, dx = -8.0 + 8.5 \] \[ \int_{-1}^{1} f(x) \, dx = 0.5 \] Therefore, the correct answer is: \[ \boxed{C} \]

EXAMPLE 7 If it is known that 6f(x) dx0 = 10 and 2f(x) dx0 = 5, find 6f(x) dx2.SOLUTION By a property of integrals, we have2f(x) dx0 + 6f(x) dx2 = 6f(x) dx0so6f(x) dx2 =  6f(x) dx0 − 2f(x) dx0 =  10 −  =  .

The graph of f is shown. Evaluate each integral by interpreting it in terms of areas.(a)    10f(x) dx0 (b)    25f(x) dx0 (c)    35f(x) dx25 (d)    45f(x) dx0

Now compare the two results.How is the definite integral   related to F(10) – F(5)?

f(x)=2x 2 +3x, find f, of, minus, 5f(−5)